Unsymmetrical Bending

Introduction:

Every section is not symmetrical about both the centroidal axes. Some sections are symmetrical only about one axis, whereas many sections as angle sections are not symmetrical about both the centroidal axes. In theory of simple bending, the section of the beam is symmetrical about the plane of bending. The simple flexural formula derived in theory of simple bending is not applicable when the section is not symmetrical about the plane of bending. In such sections, the principal axes and principal moments of inertia and the product of inertia are determined. Stresses developed in such sections of a beam are dependent on these parameters.

If the load line on a beam does not coincide with one of the principal axes of the section, the bending takes place in a plane different from the plane of principal axes. This type of bending is known as unsymmetrical bending.

The two reasons of unsymmetrical bending are as follows:1. The section is symmetrical about two axes like I-section, rectangular section, circular section but the load-line is

inclined to both the principal axes.

2. The section itself is unsymmetrical like angle section or a channel section (with vertical-web) and load-line along vertical any centroidal axes.

Figure 18.1(a) shows a beam with I-section with load-line coinciding with YY principal axis. I-section has two axes of symmetry and both these axes are the principal axes. Section is symmetrical about YY plane, i.e., the plane of bending. This type of bending is known as symmetrical bending.

Figure 18.1(b) shows a cantilever with rectangular section, which has two axes of symmetry which are principal axes but the load-line is inclined at an angle with theα  YY axis. This is the first type of unsymmetrical bending.

Then, Fig. 18.1(c) shows a cantilever with angle-section which does not have any axis of symmetry but the load-line is coinciding with the YY axis. This is the second type of unsymmetrical bending.

Figure 18.1(d) shows a channel section subjected to a vertical load passing through its centroid G. This member has been subjected to bending and twisting under the applied vertical load W. Now, the question arises; is it possible to apply the vertical load W in such a way that the channel member will bend without twisting and, if so, where the load W should be applied?

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Unsymmetrical Bending

Figure 18.1 (a) Symmetrical bending, (b) unsymmetrical bending symmetrical section but oblique load, (c) unsymmetrical bending (unsymmetrical section) and (d) unsymmetrical bending(section not symmetrical about bending plane) (e) Channel section(not symmetrical about yy axis, (f) Bending without twisting

Shear force in the flanges and web of the channel section is F1, F2 and F1, respectively, as shown in Fig. 18.1(e). Forces F1 constitute a couple F1 × h about centroid G. This couple is responsible for twisting of the member. Now, if the vertical load W or the shear force in the section is shifted from G, such that W × e = F1 × h, then the twisting couple is eliminated. So, it can be concluded that if the vertical load W, or vertical shear F is moved to the left in the channel section through a distance e, such that, F1 × h = We = Fe, the member will bend without twisting as shown in Fig.18.1(f).

Principal Axes:

Figure 18.2 shows a beam section which is symmetrical about the plane of bending Y–Y, a requirement of the theory of simple bending or symmetrical bending. G is the centroid of the section. XX and YY are the two perpendicular axes passing through the centroid. Say, the bending moment on the section (in the plane YY of the beam) is M, about the axis XX. Consider a small element of area dA with (x, y) co-ordinates.

Figure 18.2 Plane of bending yy

Stress onthe element , σ= MI xx

y (18.1)

Force on the element , dF=MydAI xx

Bendingmoment about YY axis , dM=MyxdAI xx

Total moment , M 1=∫MyxdAI xx

(18.2)

If no bending take place about YY axis, thenM1 = 0

¿∫MyxdAI xx

=0

¿ MI xx

∫ xydA=0

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¿∫ xydA=0 (18.3)

The expression  ∫ xydA  is called the product of inertia of the area about XX and YY axes, represented by Ixy. If

the product of inertia is zero about the two co-ordinate axes passing through the centroid, then the bending is symmetrical or pure bending. Such axes (about which product of inertia is zero) are called principal axes of the section and moment of inertia about the principal axes are called principal moments of inertia.

The product of inertia may be positive, negative or zero depending upon the section and co-ordinate axes. The product of inertia of a section with respect to two perpendicular axes is zero if either one of the axis is an axis of symmetry.

Parallel Axes Theorem for Product of Inertia:

Figure 18.6 shows a section with its centroid at G, and GX′ and GY′ are the two rectangular co-ordinates passing

through G. Say, the product of inertia about X′Y′ is I xy . Let us determine the product of inertia about the

axis OX and OY, i.e., Ixy.

Say, distance of G from OX axis = y  , and distance of G from OY axis = x .

Consider a small element of area dA = dx.dy

Figure 18.6

Say, co-ordinates of the element about the centroidal axis GX′, GY′ are x′, y′.Then, co-ordinates of the element about X–Y axis are,

x=x+x '∧ y= y+ y 'Therefore, the product of inertia,

I xy=∫ xydA=∫ (x+ x' ) ( y+ y ' )dxdy=∫ x ' y ' dA+ x y∫ dA+ y∫ x ' dA+¿ x∫ y 'dA ¿

¿ I xy+x y A+0+0because∫ x' dA=∫ y ' dA=0 , about centroidal axis

I xy=I xy+A x y

i.e., the product of inertia of any section with respect to any set of co-ordinate axes in its plane is equal to the product of inertia of the section with respect to the centroidal axes parallel to the co-ordinate axes plus the product of the area and the co-ordinates of the centroid of the section with respect to the given set of co-ordinate axes.

Determination of Principal Axes:

In the section ‘Introduction’, we have learnt that principal axes pass through the centroid of a section and product of inertia of the section about principal axes is zero. Figure 18.9 shows a section with centroid G and XX and XX are two co-ordinate axes passing through G. Say, UU and VV is another set of axes passing through the centroid G and inclined at an angle to the θ X–Y co-ordinate.

Consider an element of area dA at point P having co-ordinates (x, y). Say, u,v are the co-ordinates of the point P in U–V co-ordinate axes.

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So,     u = GA′ = GD + DA′ = GD + AEwhere GD = GA cosθ = x cosθAE = DA′ = y sinθor, u = x cosθ + y sinθ v = GB′ = PA′ = PE − A′E= PE − AD since A′E = AD = PA cosθ − x sinθ = y cosθ − x sinθSimilarly, x, y co-ordinates can be written in terms of u, v co-ordinates.x = GC − AC = GC − A′F = u cosθ − v sinθ(as PA′ = v and GA′ = u)y = GB = PA = AF + FP = A′C + FP = u sinθ + v cosθ

Figure 18.9 Co-ordinates along principal axesSecond moment of area about U–U axis,

I uu=∫ v2dA=∫ ( y cosθ−x sin θ )2dA=∫ y2 cos2θdA+∫ x2sin2θdA−∫2 xysinθcosθdAon simplifying,

I uu=12

( I xx+ I yy )+ 12

( I xx−I yy )cos 2θ−I xy sin 2θ (18.4)

Second moment of area about V–V,

I vv=12

( I xx+ I yy )+ 12

( I yy−I xx )c os2θ+ I xy sin 2θ (18.5)

From Eqs.(18.4) and (18.5),

Iuu + Ivv = Ixx (sin2θ + cos2θ) + Iyy (sin2θ + cos2θ) = Ixx + Iyy     (18.6)

Product of inertia about UV axes,

I uv=∫uvdA=∫ ( xcosθ+ y sinθ ) ( y cosθ−x sin θ )dA

¿∫ xy (cos2θ−sin2θ)dA+∫ y2 sinθcosθdA−∫ x2 sinθcosθdA

I uv=I xy cos2θ+ I xxsinθ2

−I yysinθ2

However, as per the condition of pure bending or symmetrical bending Iuv = 0, then U and V will be the principal axesor,                 2Ixy cos 2θ + (Ixx − Iyy)sin 2θ = 0

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tan2θ=2 I xyI yy−I xx

=I xy

( I yy−I xx )2

(18.7)

Say, θ1 and θ2 are two values of θ given by Eq. (18.7)

θ2=θ1+900

sin 2θ1=I xy

√ ( I yy−I xx )2

2+ I xy

2

∧cos2θ1=

( I yy−I xx )2

√ ( I yy−I xx )2

2+ I xy

2

Substituting these values of sin 2θ1 and cos 2θ1 in Eq. (18.4)

( I uu )θ1=12

( I xx+ I yy )+

12

( I xx−I yy ) 12

( I yy−I xx )

√[ 12 ( I yy−I xx )]2

+ I xy2

−I xy×I xy

√ 12 ( I yy−I xx )2+ I xy2

( I uu )θ1=12

( I xx+ I yy )−√[ 12 ( I yy−I xx )]2

+ I xy2 (18.8)

Similarly,

( I vv )θ1=12

( I xx+ I yy )+√[ 12 ( I yy−I xx )]2

+ I xy2 (18.9)

Now, for θ2 = θ1 + (π/2)

sin 2θ2 = sin(2θ1 + π) = −sin 2θ1

cos 2θ2 = cos(2θ1 + π) = −cos 2θ1

Substituting these values in Eqs.(18.4) and (18.5)

( I uu )θ2=12

( I xx+ I yy )+√ [ 12 ( I yy−I xx )]2

+ I xy2 (18.10)

( I vv )θ2=12

( I xx+ I yy )−√ [ 12 ( I yy−I xx )]2

+ I xy2 (18.11)

From Eqs.(18.8) to (18.11), we learn that

( I uu )θ1=( I vv )θ2 ;∧( I vv )θ1=( I uu )θ2Maximum and minimum values of Iuu and Ivv

I uu=12

( I xx+ I yy )+ 12

( I yy−I xx )cos 2θ+ I xysin 2θ

For maximum value of Iuu,

d I uudθ

=0

i .e .12¿

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tan2θ=I xy

( I yy−I xx )2

This shows that the values of (Iuu) 1θ  and (Iuu) 2θ  are the maximum and minimum values of Iuu and Ivv. These values are called the principal values of moment of inertia as Iuv = 0. The directions θ1 and θ2 are called the principal directions.

Moment of Inertia about Any Axis:

If the principal moments of inertia Iuu and Ivv are known, then moment of inertia about any axis inclined at an angle to the principal axes can be determined. Sayθ  u, v are the co-ordinates of an element of area dA in the U–

V principal axes system. X and Y are the co-ordinate axes inclined at an angle to theθ  U–V axes.

x co-ordinate of element = u cos −θ  v sin θy co-ordinate of element = u sin +θ  v cos θ

Moment of inertia,

I yy=∫ x2dA=∫ (ucosθ−vsinθ )2dA

on simplifying

I yy=I vv cos2θ+ I uusin

2θ(18.12)Similarly,

Ixx = Iuu cos2θ + Ivv sin2θ                     (18.13)

From Eqs.(18.12) and (18.13),Ixx + Iyy = Iuu + Ivv = J,where J is polar moment of inertia about an axis passing through G and normal to the section.

Stresses due to Unsymmetrical Bending:

When the load-line on a beam does not coincide with one of the principal axes of the section, unsymmetrical bending takes place. 

Figure 18.13(a) shows a rectangular section, symmetrical about XX and YY axes or with UU and VV principal axes. Load-line is inclined at an angle ø to the principal axis VV, and passing through G (centroid) or C (shear centre) of the section.

Figure 18.13Figure 18.13(b) shows an angle section which does not have any axis of symmetry. Principal axes UU and VV are inclined to axes XX and YY at an angle . Load-line is inclined at an angle ø to the vertical or at an angle (90 − ø − )θ θ to the axis UU. Load-line is passing through G(centroid of the section).

Figure 18.13(c) shows a channel section which has one axis of symmetry, i.e., XX. Therefore, UU and VV are the principal axes. G is the centroid of the section while C is the shear centre. Load-line is inclined at an angle ø to the vertical (or the axis VV) and passing through the shear centre of the section.

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Shear centre for any transverse section of a beam is the point of intersection of the bending axis and the plane of transverse section. If a load-line passes through the shear centre there will be only bending of the beam and no twisting will occur. If a section has two axes of symmetry, then shear centre coincides with the centre of gravity or centroid of the section as in the case of a rectangular, circular or I-section. For sections having one axis of symmetry only, shear centre does not coincide with centroid but lies on the axis of symmetry, as shown in the case of a channel section.

For a beam subjected to symmetrical bending only, following assumptions are made:1. The beam is initially straight and of uniform section throughout.2. Load or loads are assumed to act through the axis of bending.3. Load or loads act in a direction perpendicular to the bending axes and load-line passes through the shear centre

of transverse section.

Figure 18.14, shows the cross-section of a beam subjected to bending moment M, in the plane YY. G is the centroid of the section and XX and YY are the two co-ordinate axes passing through G. Moreover, UU and VV are the principal axes inclined at an angle to theθ  XX and YY axes, respectively. Let us determine the stresses due to bending at the point P having the co-ordinates u, v corresponding to principal axes. Moment applied in the plane YY can be resolved into two components M1 and M2.

Figure 18.14M1, moment in the plane UU = M sin θ M2, moment in the plane VV = M cos θThe components M1 and M2 have their axis along VV and UU, respectively.Resultant bending stress at the point P,

σ b=M1u

I vv+M 1 v

I uu=Msinθu

I vv+Mcosθv

Iuu=M ( v cosθI uu

+ usinθI vv

+…)(18.14)The exact nature of stress (whether tensile or compressive) depends upon the quadrant in which the point P lies. In other words sign of co-ordinates u and v is to be taken into account while determining the resultant bending stress.

The equation of the neutral axis can be determined by considering the resultant bending stress. At the neutral axis bending stress is zero, i.e.,

M ( vcosθIuu+usinθI vv )=0

¿ , v=−sin θcosθ

×IuuI vvu=−u tan α (18.15)

where tanα= sinθcosθ

I uuI vv

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This is the equation of a straight line passing through the centroid G of the section. All the points of the section on one side of the neutral axis have stresses of the same nature and all the points of the section on the other side of the neutral axis have stresses of opposite nature.

Deflection of Beams Due to Unsymmetrical Bending:

Figure 18.17 shows the transverse section of a beam with centroid G. XX and YY are two rectangular co-ordinate axes and UU and VV are the principal axes inclined at an angle to theθ  XY set of co-ordinate axes. Say the beam is subjected to a load W along the line YG. Load can be resolved into two components, i.e.,Wu = W sin θ (along UG direction)Wv = W cos θ (along VG direction)Say, deflection due to Wu is GA in the direction GU

i .e .GA=δ u=KW ul

3

E I vv

where K is a constant depending upon the end conditions of the beam and position of the load along the beam.Deflection due to Wv is GB in direction GV

i .e .GB=δ v=KW v l

3

E I uu

Total Deflection ;δ=√δu2+δ v2=K l3

E √(W u2

I vv2 +

W v2

I uu2 )= KWl3

E √( sin2θI vv2 + cos

2θI uu2 )

Figure 18.17Total deflection δ is along the direction GC, at angle γ to VV axis.

tan γ=CBGB

=GAAC

=W u

I vv×I uuW v

=W sin θW cosθ

×I uuI vv

=tan θI uuI vv

Comparing this with Eq. (18.15) of the section ‘Stresses Due to Unsymmetrical Bending’

tanα=tan θI uuI vv

where is the angle of inclination of the neutral axis with respect toα  UU axis and

tan γ=tan θI uuI vv

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where γ is the angle of inclination of direction of δ with respect to VV axis. γ = ,α  showing thereby that resultant deflection δ takes place in a direction perpendicular to the neutral axis.

Shear Centre:

Summation of the shear stresses over the section of the beam gives a set of forces which must be in equilibrium with the applied shear force F. In case of symmetrical sections such as rectangular and I-sections, the applied shear force is balanced by the set of shear forces summed over the rectangular section or over the flanges and the web of I-section and the shear centre coincides with the centroid of the section. If the applied load is not placed at the shear centre, the section twists about this point and this point is also known as centre of twist. Therefore, the shear centre of a section can be defined as a point about which the applied shear force is balanced by the set of shear forces obtained by summing the shear stresses over the section.

For unsymmetrical sections such as angle section and channel section, summation of shear stresses in each leg gives a set of forces which should be in equilibrium with the applied shear force.

Figure 18.19(a) shows an equal angle section with principal axis UU. We have learnt in previous examples that a principal axis of equal angle section passes through the centroid of the section and the corner of the equal angle as shown in the figure. Say this angle section is subjected to bending about a principal axis UU with a shearing force F at right angles to this axis. The sum of the shear stresses along the legs, gives a shear force in the direction of each leg as shown. It is obvious that the resultants of these shear forces in legs passes through the corner of the angle and unless the applied force F is applied through this point, there will be twisting of the angle section in addition to bending. This point of the equal angle section is called its shear centre or centre of twist.

For a beam of channel section subjected to loads parallel to the web, as shown in Fig. 18.19(b), the total shearing force carried by the web must be equal to applied shear force F, then in flanges there are two equal and opposite forces say F1 each. Then, for equilibrium, F ×e is equal to F1 × h and we can determine the position of the shear centre along the axis of symmetry, that is, e = (F1 × h/F).

Figure 18.19

Similarly, Fig. 18.19(c) shows a T-section and its shear centre. Vertical force in the web F is equal to the applied shear force F and horizontal forces F1 in two portions of the flange balance each other at shear centre.

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